Find the magnetic field at the point of intersection of the diagonals of a square having sides a and carrying current I.

Explanation

Distance of the point intersection (O) of the diagonals from every side, `r=(a)/(2)` [Fig. 1.95].
From the figure, angles at `O-, theta_(1)= theta_(2)=45^(@)`
Hence for each side carrying current I, magnetic field at O.
`B=(mu_(0))/(4pi)*(I)/(r)(sin theta_(1)+ sin theta_(2))`
`=(mu_(0))/(4pi)*(I)/((a)/(2))(sin 45^(@)+sin45^(@))`
`=(mu_(0))/(4pi)*(2I)/(a)*2xx(1)/(sqrt(2))=(mu_(0))/(4pi)(2sqrt(2)I)/(a)`

Now, direction of the magnetic field at O due to all four sides are same.
So, the net mangetic field at `O,4B=(mu_(0))/(4pi)(8 sqrt(2)I)/(a)`